Steiner's Hat: a Constant-Area Deltoid Associated with the Ellipse

Home , Caustic (mathematics), Deltoid curve, Envelope (mathematics), Extended side, Hypocycloid, Inellipse

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Figure 2. Left: The Negative Pedal Curve (NPC, purple) of with respect to a boundary point ′ E M is a 3-cusped (labeled Pi ) asymmetric curve (called here “Steiner’s Hat”), whose area is invariant over M, and whose asymmetric shape is aﬃnely related to the Steiner Curve [12]. ∆u(t) is the ′ instantaneous tangency point to the Hat. Right: The tangents at the cusps points Pi concur at ′ C2, the Hat’s center of area, furthermore, Pi, Pi ,C2 are collinear. Video: [10, PL#01]

As a variant to the above, we study the family of NPCs with respect to points M on the boundary of . As shown in Figure 2, this yields a family of asymmetric, constant-area 3-cuspedE deltoids. We call these curves “Steiner’s Hat” (or ∆), since under a varying aﬃne transformation, they are the image of the Steiner Curve (aka. Hypocycloid), Figure 3. Besides these remarks, we’ve observed: Main Results:

The triangle T ′ defined by the 3 cusps Pi′ has invariant area over M, Fig- • ure 7. The triangle T defined by the pre-images P of the 3 cusps has invariant area • i over M, Figure 7. The Pi are the 3 points on such that the corresponding tangent to the envelope is at a cusp. E The T are a Poncelet family with fixed barycenter; their caustic is half the • size of , Figure 7. E Let C2 be the center of area of ∆. Then M, C2, P1, P2, P3 are concyclic, • Figure 7. The lines P C are tangents at the cusps. i − 2 Each of the 3 circles passing through M, P , P ′, i = 1, 2, 3, osculate at • i i E Pi, Figure 8. Their centers define an area-invariant triangle T ′′ which is a half-size homothety of T ′. The paper is organized as follows. In Section 3 we prove the main results. In Sections 4 and 5 we describe properties of the triangles defined by the cusps and their pre-images, respectively. In Section 6 we analyze the locus of the cusps. In Section 6.1 we characterize the tangencies and intersections of Steiner’s Hat with the ellipse. In Section 7 we describe properties of 3 circles which osculate the ellipse at the cusp pre-images and pass through M. In Section 8 we describe relationships between the (constant-area) triangles with vertices at (i) cusps, (ii) cusp pre-images, and (iii) centers of osculating circles. In Section 9 we analyze a fixed-area deltoid obtained from a “rotated” negative pedal curve. The paper concludes in Section 10 with a table of illustrative videos. Appendix A provides STEINER’S HAT: A CONSTANT-AREA DELTOID 3

Figure 3. Two systems which generate the 3-cusp Steiner Curve (red), see [2] for more methods. Left: The locus of a point on the boundary of a circle of radius 1 rolling inside another of radius 3. Right: The envelope of Simson Lines (blue) of a triangle T (black) with respect to points P (t) on the Circumcircle [12]. Q(t) denotes the corresponding tangent. Nice properties include (i) the area of the Deltoid is half that of the Circumcircle, and (ii) the 9-point circle of T (dashed green) centered on X5 (whose radius is half that of the Circumcircle) is internally tangent to the Deltoid [13, p.231]. explicit coordinates for cusps, pre-images, and osculating circle centers. Finally, Appendix B lists all symbols used in the paper.

2. Preliminaries Let the ellipse be deﬁned implicitly as: E x2 y2 (x, y)= + 1=0, c2 = a2 b2 E a2 b2 − − where a>b> 0 are the semi-axes. Let a point P (t) on its boundary be parametrized as P (t) = (a cos t,b sin t). Let P = (x ,y ) R2. Consider the family of lines L(t) passing through P (t) 0 0 0 ∈ and orthogonal to P (t) P0. Its envelope ∆ is called antipedal or negative pedal curve of . − ConsiderE the spatial curve deﬁned by

(P )= (x,y,t): L(t,x,y)= L′(t,x,y)=0 . L 0 { } The projection (P0) = π( (P0)) is the envelope. Here π(x,y,t) = (x, y). In general, (P ) isE regular, butL (P ) is a curve with singularities and/or cusps. L 0 E 0 Lemma 1. The envelope of the family of lines L(t) is given by: 1 x(t)= [(ay sin t ab)x by2 cos t c2y sin(2t) w 0 − 0 − 0 − 0 b + ((5a2 b2)cos t c2 cos(3t))] 4 − − 1 y(t)= [ ax2 sin t + (by cos t + c2 sin(2t))x aby w − 0 0 0 − 0 a (1) ((5a2 b2) sin t c2 sin(3t)] − 4 − − 4 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN where w = ab bx cos t ay sin t. − 0 − 0 Proof. The line L(t) is given by: (x a cos t)x + (y b sin t)y + a2 cos2 t + b2 sin2 t ax cos t by sin t =0 0 − 0 − − 0 − 0 Solving the linear system L(t)= L′(t)=0 in the variables x, y leads to the result.

Triangle centers will be identifed below as Xk following Kimberling’s Encyclope- dia [6], e.g., X1 is the Incenter, X2 Barycenter, etc.

3. Main Results

Proposition 1. The NPC with respect to Mu = (a cos u,b sin u) a boundary point of is a 3-cusp closed curve given by ∆ (t) = (x (t),y (t)), where E u u u 1 x (t)= c2(1 + cos(t + u))cos t a2 cos u u a − 1 (2) y (t)= c2 cos t sin(t + u) c2 sin t a2 sin u u b − −

Proof. It is direct consequence of Lemma 1 with P0 = Mu.

Expressions for the 3 cusps Pi′ in terms of u appear in Appendix A. Remark 1. As a/b 1 the ellipse becomes a circle and ∆ shrinks to a point on the boundary of said circle.→

Remark 2. Though ∆ can never have three-fold symmetry, for Mu at any ellipse vertex, it has axial symmetry.

Remark 3. The average coordinates C¯ = [¯x(u), y¯(u)] of ∆u w.r.t. this parametriza- tion are given by:

1 2π (a2 + b2) x¯(u)= x (t) dt = cos u 2π u − 2a Z0 1 2π (a2 + b2) (3) y¯(u)= y (t) dt = sin u 2π u − 2b Z0 Theorem 1. ∆u is the image of the 3-cusp Steiner Hypocycloid under a varying aﬃne transformation. S Proof. Consider the following transformations in R2:

cos u sin u x rotation: Ru(x, y)= sin u cos u y − ! ! translation: U(x, y) = (x, y)+ C¯ 1 2 2 homothety: D(x, y)= 2 (a b )(x, y). x y − linear map: V (x, y) = ( a , b ) The hypocycloid of Steiner is given by (t) = 2(cos t, sin t)+(cos2t, sin 2t) [7]. Then: S − STEINER’S HAT: A CONSTANT-AREA DELTOID 5

(4) ∆ (t) = [x (t),y (t)] = (U V D R ) (t) u u u ◦ ◦ ◦ u S Thus, Steiner’s Hat is of degree 4 and of class 3 (i.e., degree of its dual).

Corollary 1. The area of A(∆) of Steiner’s Hat is invariant over Mu and is given by:

(a2 b2)2π c4π (5) A(∆) = − = 2ab 2ab Proof. The area of (t) is xdy =2π. The Jacobian of (U S D R ) given by S S ◦ ◦ ◦ u Equation 4 is constant and equal to c4/4ab. R Noting that the area of is πab, Table 1 shows the aspect ratios a/b of required to produce special area ratios.E E

a/b approx. a/b A(∆)/A( ) E 2+ √3 1.93185 1 ϕ = (1+p √5)/2 1.61803 1/2 √2 1.41421 1/4 1 1 0

Table 1. Aspect ratios yielding special area ratios of main ellipse to Steiner’s Hat ∆. E

It is well known that if M is interior to then the NPC is a 2-cusp or 4-cusp curve with one or two self-intersections. E Remark 4. It can be shown that when M is interior to the iso-curves of signed area of the NPC are closed algebraic curves of degree 10,E concentric with and symmetric about both axes, see Figure 4. E It is remarkable than when M moves from the interior to the boundary of , the iso-curves transition from a degree-10 curve to a simple conic. E Remark 5. It can also be shown that when M is exterior to , the NPC is a two-branched open curve, see Figure 5. E

Proposition 2. Let C2 be the center of area of ∆u. Then C2 = C¯. Proof. The center of area is defined by 1 C = x dx dy, y dx dy . 2 A(∆) Zint(∆) Zint(∆) ! Using Green’s Theorem, evaluate the above using the parametric in Equation (1). This yields the expression for C¯ in Equation 3. Alternatively, one can obtain the same result from the affine transformation defined in Theorem 1. Referring to Figure 6(left): Corollary 2. The locus of C is an ellipse always exterior to a copy of rotated 2 E 90◦ about O. 6 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

Figure 4. The isocurves of signed area for the negative pedal curve when M is interior to the ellipse are closed algebraic curves of degree 10. These are shown in gray for an NPC with two cusps (left), and 4 cusps (right).

Figure 5. Left: When M is exterior to the NPC is a two-branched open curve. One branch is smooth and non-self-intersecting, and theE other has 2 cusps and one self-intersection. Right: Let t1, t2 be the parameters for which MP (t) is tangent to . At these positions, the NPC intersects E the line at inﬁnity in the direction of the normal at P (t1), P (t2), i.e., the lines through P (t) perpendicular to P (t) M are asymptotes. −

Proof. Equation 3 describes an ellipse. Since a2 + b2 2ab the claim follows directly. ≥

Let T denote the triangle of the pre-images P on of the Hat’s cusps, i.e. P (t ) i E i such that ∆uti is a cuspid. Explicit expressions for the Pi appear in Appendix A. Referring to Figure 7:

Theorem 2. The points M, C2, P1, P2, P3 are concyclic. STEINER’S HAT: A CONSTANT-AREA DELTOID 7

′ Figure 6. Left: The area center C2 of Steiner’s Hat coincides with the barycenter X2 of the ′ ′ (dashed) triangle T deﬁned by the cusps. Over all M, both the Hat and T have invariant area. ◦ C2’s locus (dashed purple) is elliptic and exterior to a copy of rotated 90 about its center (dashed ′ E black). Right: Let Pi (resp. Pi), i = 1, 2, 3 denote the Hat’s cusps (resp. their pre-images on ), ′ E colored by i. Lines PiPi concur at C2.

u u 2π u 4π Proof. ∆u(t) is singular at t1 = 3 , t2 = 3 3 and t3 = 3 3 . Let P = [a cos t ,b sin t ],i =1, 2, 3. The− circle passing− − through these− is given− by: i i i K c2 cos u c2 sin u 1 (6) (x, y)= x2 + y2 x + y (a2 + b2)=0. K − 2a 2b − 2 Also, (M)= (a cos u,b sin u)=0. K K The center of is (M + C2)/2. It follows that C2 and that MC2 is a diameter of . K ∈ K K In 1846, Jakob Steiner stated that given a point M on an ellipse , there exist 3 other points on it such that the osculating circles at these points passE through M [8, page 317]. This property is also mentioned in [4, page 97, Figure 3.26]. It turns out the cusp pre-images are said special points! Referring to Figure 8:

Proposition 3. Each of the 3 circles through M, P , P ′, i =1, 2, 3, osculate Ki i i E at Pi.

Proof. The circle passing through M, P and P ′ is given by K1 1 1 u u (x, y) =2 ab(x2 + y2) 4bc2 cos3 x 4ac2 sin3 y K1 − 3 − 3 2u +ab 3c2 cos a2 b2 =0. 3 − − 8 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

Figure 7. The cusp pre-images Pi deﬁne a triangle T (orange) whose area is invariant over M. Its barycenter X2 is stationary at the center of , rendering the latter its Steiner Ellipse. Let C2 E denote the center of area of Steiner’s Hat. The 5 points M,C2, P1, P2, P3 lie on a circle (orange), with center at X3 (circumcenter of T ). Over all M, the T are a constant-area Poncelet family inscribed on and tangent to a concentric, axis-aligned elliptic caustic (dashed orange), half the size of , i.e.,E the latter is the (stationary) Steiner Inellipse of the T . Note also that M is the E Steiner Point X99 of T since it is the intersection of its Circumcircle with the Steiner Ellipse. Furthermore, the Tarry Point X98 of T coincides with C2, since it is the antipode of M = X99 [6]. Video: [10, PL#02,#05].

Recall a circle osculates an ellipse if its center lies on the evolute of said ellipse, given by [4]:

c2 cos3 t c2 sin3 t (7) ∗(t)= , E a − b u It is straightforward to verify that the center of 1 is P1′′ = ∗( 3 ). A similar analysis can be made for and . K E − K2 K3 4 3πc Since the area of ∗ is A( ∗)= , and the area of ∆ is given in Equation 5: E E 8ab

Remark 6. The area ratio of ∆ and the interior of ∗ is equal to 4/3. , E STEINER’S HAT: A CONSTANT-AREA DELTOID 9

′ Figure 8. The circles passing through a cusp Pi , its pre-image Pi, and M osculate at the Pi. ′′ ′′ E The centers Pi of said circles deﬁne a triangle T (dashed black) whose area is constant for all ′′ M. X2 denotes its (moving) barycenter. Video: [10, PL#03,#05].

3.1. Why is ∆ aﬃne to Steiner’s Curve. Up to projective transformations, there is only one irreducible curve of degree 4 with 3 cusps. In a projective coordinate frame (x0 : x1 : x2) with the cusps as base points (1:0:0), (0:1:0) and (0:0:1) and the common point of the cusps’ tangents as unit point (1:1:1), the quartic has the equation

x2x2 + x2x2 + x2x2 2x x x (x + x + x )=0 0 1 0 2 1 2 − 0 1 2 0 1 2 At Steiner’s three-cusped curve, the cusps form a regular triangle with the tangents passing through the center. Hence, whenever a three-cusped quartic has the meeting point of the cusps’ tangents at the center of gravity of the cusps, it is aﬃne to Steiner’s curve, since there is an aﬃne transformation sending the four points into a regular triangle and its center.

4. The Cusp Triangle

Recall T ′ = P1′P2′P3′ is the triangle deﬁned by the 3 cusps of ∆.

Proposition 4. The area A′ of the cusp triangle T ′ is invariant over M and is given by: 27√3 c4 A′ = 16 ab Proof. The determinant of the Jacobian of the aﬃne transformation in Theorem 1 4 c is J = . Therefore, the area of T ′ is simply J A , where A is the area of an | | 4ab | | e e equilateral triangle inscribed in a circle of radius 3 with side 3√3. Referring to Figure 6: 10 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

′ ′ ′ ◦ Figure 9. Left: The Steiner Ellipse of triangle T deﬁned by the Pi is a scaled-up and 90 - E ′ rotated copy of . Right: At a/b = (1+ √10)/3 1.38743, and have the same area. E ≃ E E

Proposition 5. The barycenter X2′ of T ′ coincides with the center of area C2 of ∆. Proof. Direct calculations yield X2′ = C2.

Referring to Figure 9:

Proposition 6. The Steiner Ellipse ′ of T ′ has constant area and is a scaled E version of rotated 90◦ about O. E

Proof. ′ passes through the vertices of T ′ and is centered on C2 = X2′ . Direct calculationsE yield the following implicit equation for it:

a2x2 + b2y2 + a2 + b2 (a cos ux + b sin uy) a2 2 b2 2 a2 b2 =0 − − − 2 2 3c 3c Its semi-axes are b′ = 2a and a′ = 2b . Therefore ′ is similar to a 90◦-rotated copy of . E E

Remark 7. This proves that T ′ can never be regular and ∆ has never a three-fold symmetry.

4 9 c Corollary 3. The ratio of area of ′ and is given by 2 2 , and at a/b = E E 4 a b (1 + √10)/3, the two ellipses are congruent.

Proposition 7. The Steiner Point X99′ of T ′ is given by:

a2 2 b2 2 a2 b2 X99′ (u)= − cos u, − sin u " a − b # STEINER’S HAT: A CONSTANT-AREA DELTOID 11

Proof. By deﬁnition X is the intersection of the circumcircle of T ( ) with the 99 K Steiner ellipse. The Circumcircle ′ of the triangle T ′ = P ′, P ′, P ′ is given by: K { 1 2 3} 2 2 2 2 ′(x, y) =8 a b x + y K +2 a cos u 3 a4 2 a2b2 +7 b4 x +2 b sin u 7 a4 2 a2b2 +3 b4 y − − a2 + b2 c2 a2 + b2 cos2 u +5 a4 14 a2b2 +5 b4 =0. − − With the above, straightforward calculations lead to the coordinates of X99′ .

5. The Triangle of Cusp Pre-Images Recall T = P P P is the triangle deﬁned by pre-images on to each cusp of ∆. 1 2 3 E Proposition 8. The barycenter X2 of T is stationary at O, i.e., is is Steiner Ellipse [12]. E Proof. The triangle T is an aﬃne image of an equilateral triangle with center at 0 and P = (t )= ( u (i 1) 2π ). So the result follows. i E i E − 3 − − 3

Remark 8. M is the Steiner Point X99 of T . Proposition 9. Over all M, the T are an N = 3 Poncelet family with external conic with the Steiner Inellipse of T as its caustic [12]. Futhermore the area of E 3√3ab these triangles is invariant and equal to 4 . Proof. The pair of concentric circles of radius 1 and 1/2 is associated with a Poncelet 1d family of equilaterals. The image of this family by the map (x, y) (ax,by) produces the original pair of ellipses, with the stated area. Alternatively,→ the ratio of areas of a triangle to its Steiner Ellipse is known to be 3√3/(4π) [12, Steiner Circumellipse] which yields the area result.

6. Locus of the Cusps

We analyze the motion of the cusps Pi′ of Steiner’s Hat ∆ with respect to con- tinuous revolutions of M on . Referring to Figure 10: E Remark 9. The locus (u) of the cusps of ∆ is parametrized by: C

3c2 1 u 1 u a2 + b2 1 1 (8) (u): cos , sin cos u, sin u C 2 a 3 b 3 − 2 a b This is a curve of degree 6, with the following implicit equation:

6 6 6 6 2 2 2 2 2 2 2 2 4 a x 4 b y 12 a x b y a x + b y − − − 4 4 2 2 4 4 4 4 2 2 4 4 2 2 4 2 2 4 2 2 +12 a a a b + b x + 12 b a a b + b y + 24 a b a a b + b x y − − − 2 2 2 2 2 4 2 2 4 2 3 a 2 a b a + b 2 a 5 a b + 5 b x − − − 2 2 2 2 2 4 2 2 4 2 +3 b a 2 b a + b 5 a 5 a b + 2 b y − − 2 2 2 2 2 2 2 2 2 + 2 a b a 2 b a + b = 0 − − 12 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

Proposition 10. It can be shown that over one revolution of M about , C2 will cross the ellipse on four locations W , j =1, , 4 given by: E j ···

1 2 2 2 2 Wj = a a +3b , b 3a + b 2√a2 + b2 ± ± p p At each such crossing, C2 coincides with one of the pre-images.

Proof. From the coordinates of C2 given in equation (3) in terms of the parameter u, one can derive an equation that is a necessary and suﬃcient condition for C2 to happen, by substituting those coordinates in the ellipse equation x2/a2+y2/b2∈E1= − 0. Solving for sin u and substituting back in the coordinates of C2, one easily gets the four solutions W1, W2, W3, W4. Now, assume that C . The points M, P , P , P , C must all be in both the 2 ∈E 1 2 3 2 ellipse and the circumcircle of P1P2P3. Since the two conics have at most 4 intersectionsE (counting multiplicity),K 2 of those 5 points must coincide. It is easy to verify from the previously-computed coordinates that M can only coincide with the preimages P1, P2, P3 at the vertices of . In such cases, owing to the symmetry of the geometry about either the x- or y-axis,E the circle must be tangent to at M. Thus, that intersection will count with multiplicityK (of at least) 2, so anotherE pair of those 5 points must also coincide. Since P1, P2, P3 must all be distinct, C2 will coincide with one of the preimages. However, this can never happen, since if M is on one of the vertices of , C won’t be in . In any other case, since P , P , P E 2 E 1 2 3 must be distinct and C2 is diametrically opposed to M in , C2 must coincide with one of the preimages. K

Remark 10. C2 will visit each of the preimages cyclically. Moreover, upon 3 revolutions (with 12 crossings in total), each Pi will have been visited four times and the process repeats.

Figure 10. The loci of the cusps of ∆ (dashed line) is a degree-6 curve with 2 internal lobes with either 2, 3, or 4 self-intersections. From left to right, a/b = 1.27, √2, 1.56 . Note that at { } a/b = √2 the two lobes touch, i.e., the cusps pass through the center of . Also shown is the E elliptic locus of C2 (purple). Points Zi (resp. Wi) mark oﬀ the intersection of the locus of the cusps (resp. of C2) with . These never coincide Video: [10, PL#04]. E

6.1. Tangencies and Intersections of the Deltoid with the Ellipse. STEINER’S HAT: A CONSTANT-AREA DELTOID 13

Deﬁnition 1 (Apollonius Hyperbola). Let M be a point on an ellipse with semi-axes a,b. Consider a hyperbola , known as the Apollonius HyperbolaE of M [5]: H

: (x, y) M, (y/b2, x/a2) =0. H h − − i Notice that for P on , only the points for which the normal at P points to M will lie on . See also [4E, page 403]. Additionally,H passes through M and O, and its asymptotes are parallel to the axes of . H E Proposition 11. ∆ is tangent to at , at 1, 2 or 3 points Q depending on E E ∩ H i whether M is exterior or interior to the evolute ∗. E Proof. ∆ is tangent to at some Q if the normal of at Q points to M = E i E i (Mx,My), i.e., when intersects with . It can be shown that their x coordinate is given by the real rootsH of: E

(9) (x)= c4x3 c2M (a2 + b2)x2 a4(a2 2b2)x + a6M =0 Q − x − − x The discriminant of the above is:

4c4a6(a2 M 2)[(a2 b2)(a2 + b2)3M 2 + a4(a2 2b2)3]. − − x − x − 2 2 2 2 3 2 4 2 2 3 Let x∗ denote the solutions to (a b )(a + b ) M + a (a 2b ) =0. As- ± − x − suming a>b, Equation 9 has three real solutions when x < x∗. The intersections | | of the evolute ∗ with the ellipse are given by the four points ( x∗, y∗), where: E E ± ±

3 3 a2√a4 b4 a2 2 b2 2 b2√a4 b4 2 a2 b2 2 (10) x∗ = − − , y∗ = − − (a4 b4) (a2 + b2) (a4 b4) (a2 + b2) − −

For M ∗ two coinciding roots result in a 4-point contact between ∆ and the ellipse. ∈E∩E

Let M = (Mx,My) be a point on and (x) denote the following cubic poly- nomial: E J

(11) (x) = (a2 + b2)2x2 2M c2(a2 + b2)x 4a4b2 + M 2(a2 + b2)2 J − x − x

Proposition 12. ∆ intersects at (x) (x) =0, in at least 3 and up to 5 locations locations, where is asE in EquationQ J 9. Q Proof. As before, M = (M ,M ) = (a cos u,b sin u) and P = (x, y) = (a cos t,b sin t). x y ∈E The intersection ∆u(t) with is obtained by setting (∆u(t)) = 0. Using Equa- tion 1, obtain the following system:E E 14 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

Figure 11. Steiner’s Hat ∆ (purple, top cusp not shown) is tangent to at the intersections E Qi of the Apollonius Hyperbola (olive green) with , excluding M. Notice passes through ∗ the center of . Left: When M His exterior to the evoluteE (dashed purple),H only one tangent E ∗ E Q1 is present. Right: When M is interior to , three tangent points Qi, i = 1, 2, 3 arise. The ∗ ∗ intersections of are given in Equation 10. Note:E the area ratio of ∆-to- is always 4/3. E E

F (x, y)=b2 a2 2 M 2 a2 + b2 c4x4 +2 a2M M a2 + b2 c4x3y − x x y +2 a2b2M c2 a4 + b4 x3 2 a4M c2 a4 + b4 x2y x − y + a4b2 a2 + b2 3 a4 4 a2b2 +2 b4 + a2b2c2M 2 3 a2 b2 a2 + b2 x2 − − x − 2 a6M M c2 a2 + b2 xy 2 b2M a6 a4 a2b2 + b4 x − x y − x − +2 a12M y + a8b2 2 a4 (a2 + b2)M 2 =0 y − x x2 y2 (x, y)= + 1=0 E a2 b2 − By Bézout’s theorem, the system (x, y)= F (x, y)=0 has eight solutions, with algebraic multiplicities taken into account.E ∆ has three points of tangency with ellipse, some of which may be complex, which by Proposition 11 are given by the zeros of (x). Eliminating y by computing the resultant we obtain an Equation G(x)=0Qof degree 8 over x. Manipulation with a Computer Algebra System yields a compact representation for G(x): G(x)= (x)2 (x) Q J with as in Equation 11. If Mx a, the solutions of (x)=0 are real and given J 2 2 | 2 |≤2 2 J by Jx = [c Mx 2ab a Mx ]/(a + b ) and Jx a. ± − | |≤ p Referring to Figure 10:

Proposition 13. When a cusp P ′ crosses the boundary of , it coincides with its i E pre-image P at Z = 1 ( a2, b2). i i √a2+b2 ± ±

Proof. Assume Pi′ is on . Since Pi′ is on the circle i deﬁned by M, Pi and Pi′ which osculates at P ,E this circle intersects at P Kwith order of contact 3 or 4. E i E i STEINER’S HAT: A CONSTANT-AREA DELTOID 15

By construction, we have MP P P ′, so M and P ′ are diametrically opposite in i ⊥ i i i . Thus, M and P ′ must be distinct. Since two conics have at most 4 intersections Ki i (counting multiplicities), we either have Pi′ = Pi or Pi = M. The second case will only happen when M is on one of the four vertices of the ellipse , in which case E the osculating circle has order of contact 4, so P ′ could not also be in the ellipse Ki i in the ﬁrst place. Thus, Pi′ = Pi as we wanted. Substituting the parameterization of P in the equation of , we explicitly ﬁnd i E the four points Z at which P ′ can intersect the ellipse . i i E

7. A Triad of Osculating Circles Recall are the circles which osculate at the pre-images P , see Figure 8. De- Ki E i ﬁne a triangle T ′′ by the centers Pi′′ of the i. These are given explicit coordinates in Appendix A. Referring to Figure 12: K

Proposition 14. Triangles T ′ and T ′′ are homothetic at ratio 2:1, with M as the homothety center.

Proof. From the construction of ∆, for each i = 1, 2, 3 we have MP P P ′, that i ⊥ i i is, ∠MP P ′ = 90◦. Hence, MP ′ is a diameter of the osculating circle that goes i i i Ki through M, P , and P ′ as proved in Proposition 3. Thus, the center P ′′ of is the i i i Ki midpoint of MPi′ and therefore Pi′ is the image of Pi′′ under a homothety of center M and ratio 2.

Corollary 4. The area A′′ of T ′′ is invariant over all M and is 1/4 that of T ′.

Proof. This follows from the homothety, and the fact that the area of T ′ is invariant from Proposition 4.

Proposition 15. Each (extended) side of T ′ passes through an intersection of two osculating circles. Moreover, those sides are perpendicular to the radical axis of said circle pairs.

Proof. It suﬃces to prove it for one of the sides of T ′ and the others are analogous. Let M1 be the intersection of 2 and 3 diﬀerent than M and let M1/2 be the midpoint of M and M . SinceKMM isK the radical axis of and , the lines 1 1 K2 K3 P2′′P3′′ and MM1 are perpendicular and their intersection is M1/2. Applying an homothety of center M and ratio 2, we get that the lines P2′P3′ and MM1 (the radical axis) are perpendicular and their intersection is M1, as desired.

Corollary 5. The Steiner ellipse ′′ of triangle T ′′ is similar to ′. In fact, ′′ = 1 E E E ′. 2 E Proof. This follows from the homothety of T ′ and T ′′.

8. Relations between T,T ′,T ′′

As before we identify Triangle centers as Xk after Kimberling’s Encyclopedia [6].

Proposition 16. The circumcenter X3 of T coincides with the barycenter X2′′ of T ′′. 16 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

′ ′′ Figure 12. Lines connecting each cusp Pi to the center Pi of the circle which osculates at E the pre-image Pi concur at M. Note these lines are diameters of said circles. Therefore M is the ′ ′′ ′′ perspector of T and T , i.e., the ratio of their areas is 4. This perspectivity implies C2, X2 , M ′′ are collinear. Surprisingly, the X2 coincides with the circumcenter X3 of the pre-image triangle T (not drawn).

Proof. Follows from direct calculations using the coordinate expressions of Pi and Pi′′. In fact, c2 cos u sin u X′′ = X = , . 2 3 4 a − b

Corollary 6. The homothety with center M and factor 2 sends X2′′ to X2′ = C2.

Proposition 17. The lines joining a cusp Pi′ to its preimage Pi concur at ∆’s center of area C2.

Proof. From Theorem 2, points M and C2 both lie on the circumcircle of T and form a diameter of this circle. Thus, for each i = 1, 2, 3, we have ∠MPiC2 = 90◦. ∠ By construction, MPiPi′ = 90◦, so Pi, Pi′, and C2 are collinear as desired. Referring to Figure 14(left): STEINER’S HAT: A CONSTANT-AREA DELTOID 17

′ ′′ Figure 13. Consider a reference triangle (blue), and its pedal (red) and antipedal (green) triangles with respect to some point Z. ConstructionT lines for bothT pedal and antipedalT (dashed ′ ′′ red, dashed green) imply that Z is an orthology center simultaneousy for both , and , , ′ ′′ T T T T i.e. these pairs are orthologic. Also shown are H and H , the 2nd orthology centers of said pairs (construction lines omitted). Non-transitivity arises from the fact that perpendiculars dropped ′ ′′ from the vertices of to the sides of (dashed purple, feet are marked X) are non-concurrent T T ′ ′′ (purple diamonds mark the three disjoint intersections), i.e., , are not orthologic. T T

Corollary 7. C2 = X2′ is the perspector of T ′ and T . Lemma 2. Given a triangle , and its Steiner Ellipse Σ, the normals at each vertex pass through the OrthocenterT of , i.e., they are the altitudes. T Proof. This stems from the fact that the tangent to Σ at a vertex of is parallel to opposide side of [12, Steiner Circumellipse]. T T Referring to Figure 14(right):

Proposition 18. The orthocenter X4 is the perspector of T and T ′′. Equivalently, a line connecting a vertex of T to the respective vertex of T ′′ is perpendicular to the opposite side of T . Proof. Since T has fixed X , is its Steiner Ellipse. The normals to the latter at 2 E Pi pass through centers Pi′′ since these are osculating circles. So by Lemma 2 the proof follows. Definition 2. According to J. Steiner [3, p. 55], two triangles ABC and DEF are said to be orthologic if the perpendiculars from A to EF , from B to DF , and from C to DE are concurrent. Furthermore, if this holds, then the perpendiculars from D to BC, from E to AC, and from F to AB are also concurrent. Those two points of concurrence are called the centers of orthology of the two triangles [9]. Note that orthology is symmetric but not transitive [9, p. 37], see Figure 13 for a non-transitive example involving a reference, pedal, and antipedal triangles. Lemma 3. Let P denote the reflection of P about O = X for i =1, 2, 3. Then − i i 2 the line from M to P1 is perpendicular to the line P2′′P3′′, and analogously for P2, and P . − − − 3 18 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

′ Figure 14. Left: T and T are perspective on X2. They are also orthologic, with orthology ′′ centers X671 and the reﬂection of X99 on X4. Right: T and T are perspective on X4. They are also orthologic, with orthology centers X4 and X671 .

Proof. This follows directly from the coordinate expressions for points M, P = (t ) i E i and P ′′ = ∗(t ). It follows that M + P , P ′′ P ′′ =0. i E i h 1 2 − 3 i Referring to Figure 14(right):

Theorem 3. Triangles T and T ′ are orthologic and their centers of orthology are the reﬂections X671 of M on X2 and on X4.

Proof. We denote by X671 the reflection of M = X99 on O = X2. From Lemma 3, the line through M and P is perpendicular to P ′′P ′′. Reflecting about O = X , − 1 2 3 2 the line P X is also perpendicular to P ′′P ′′. Since P ′′P ′′ P ′P ′ from the 1 671 2 3 2 3 k 2 3 homothety, we get that P X P ′P ′. This means the perpendicular from P to 1 671 ⊥ 2 3 1 P2′P3′ passes through X671. Analogously, the perpendiculars from P2 to P1′P3′ and from P3 to P1′P2′ also go through X671. Therefore T and T ′ are orthologic and X671 is one of their two orthology centers. Let Xh be the reflection of M on X4. From Proposition 18, the line through P1 and P1′′ passes through the orthocenter X4 of T , that is, P1′′ is on the P1-altitude of T . This means that the perpendicular from P1′′ to the line P2P3 passes through X4. Applying the homothety with center M and ratio 2, the perpendicular from X1′ to X2X3 passes through Xh. Analogously, the perpendiculars from X2′ to X1X3 and from X3′ to X1X2 also pass through Xh. Hence, Xh is the second orthology center of T and T ′.

Theorem 4. Triangles T and T ′′ are orthologic and their centers of orthology are X4 and the reﬂection X671 of M on X2.

Proof. From Proposition 18, the perpendiculars from P1′′ to P2P3, from P2′′ to P1P3, and from P3′′ to P1P2 all pass through X4. Thus, triangles T and T ′′ are orthologic and X4 is one of their two orthology centers. As before, we denote by X671 the reﬂection of M = X99 on O = X2. Again, from Lemma 3, the line through M and P is perpendicular to P ′′P ′′, so reﬂecting it − 1 2 3 at X , we get that P X P ′′P ′′. Since the triangles T ′′ and T ′ have parallel 2 1 671 ⊥ 2 3 sides, we get P X P ′P ′ P ′′P ′′. Thus, X is the second orthology center of 1 671 ⊥ 1 3 k 1 3 671 T and T ′′. STEINER’S HAT: A CONSTANT-AREA DELTOID 19

′′ Figure 15. The perspectrix of T, T is perpendicular to the line through X4 and X671. Compare ′ with Figure 13: the perspectrix of T, T is perpendicular to the Euler Line of T .

Theorem 5 (Sondat’s Theorem). If two triangles are both perspective and orthologic, their centers of orthology and perspectivity are collinear. Moreover, the line through these centers is perpendicular to the perspectrix of the two triangles [11, 9].

Referring to Figure 15:

Theorem 6. The perspectrix of T and T ′ is perpendicular to the Euler Line of T .

Proof. Since T and T ′ are both orthologic and perspective from Corollary 7 and Theorem 3, by Sondat’s Theorem, their perspectrix is perpendicular to the line through their orthology centers (reﬂections of M at X2 and X4) and perspector (X2′ = C2 = X98 =reﬂection of M at X3). By applying a homothety of center M and ratio 1/2, this last line is parallel to the line through X2, X3, and X4, the Euler line of T . Therefore the perspectrix of T and T ′ is perpendicular to the Euler line of T .

Proposition 19. The perspectrix of T and T ′′ is perpendicular to the line X4X671 (which is parallel to the line through M and X376, the reﬂection of X2 at X3).

Proof. Since T and T ′′ are both orthologic and perspective from Proposition 18 and Theorem 4, by Sondat’s Theorem, their perspectrix is perpendicular to the line through their orthology centers X4 and X671. Reflecting this last line at X2, we find that it is parallel to the line through M and the reflection of X4 at X2, which is the same as the reflection of X2 at X3.

Table 2 lists a few pairs of triangle centers numerically found to be common over T,T ′ or T,T ′′. 20 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

Figure 16. From left to right: for a ﬁxed M, the line passing through P (t) and perpendicular to the segment P (t) M (dashed) purple is rotated clockwise by θ = 30, 45, 60 degrees, respectively ∗ (dashed olive green).− For all P (t) these envelop new constant-area crooked hats ∆ (olive green) 2 whose areas are cos(θ) that of ∆. For the θ shown, these amount to 3/4, 1/2, 1/4 of the area of ∆ ∗ ∗ (purple). As one varies θ, the center of area C2 of ∆ sweeps the circular arc between C2 and M with center at angle 2θ. The same holds for the cusps running along the corresponding osculating circles (shown dashed red, green, blue), which are stationary and independent of θ. Video: [10, PL#06]

T T ′ T ′′

X3 – X2′′

X4 – X671′′

X5 – X115′′

X20 – X99′′

X76 – X598′′

X98 X2′ –

X114 X230′ –

X382 – X148′′

X548 – X620′′

X550 – X2482′′

′ ′′ Table 2. Triangle Centers which coincide T, T or T, T .

9. Addendum: Rotated Negative Pedal Curve The Negative Pedal Curve is the envelope of lines L(t) passing through P (t) and perpendicular to P (t) M. Here we consider the envelope ∆θ∗ of the L(t) rotated clockwise a ﬁxed θ about− P (t); see Figure 16.

Proposition 20. ∆θ∗ is the image of the NPC ∆ under the similarity which is the product of a rotation about M through θ and a homothety with center M and factor cos θ.

Proof. For variable parameter t, the lines L(t) and Lθ∗(t) perform a motion which sends P along , while the line through P orthogonal to L(t) slides through the ﬁxed point M. DueE to basic results of planar kinematics [1, p. 274], the instantaneous center of rotation I lies on the normal to at P and on the normal to MP at E STEINER’S HAT: A CONSTANT-AREA DELTOID 21

C* C I

P(t) M

∗ ∗ Figure 17. The construction of points C, C of the envelopes ∆ (red) and ∆ (blue) with the help of the instant center of rotation I reveals that the rotation about M through θ and scaling with ∗ factor cos θ sends ∆ to ∆ (Proposition 20).

M. We obtain a rectangle with vertices P , M and I. The fourth vertex is the enveloping point C of L(t). The enveloping point C∗ of Lθ∗ is the pedal point of I. Since the circumcircle of the rectangle with diameter MC also passes through C∗, we see that C∗ is the image of C under the stated similarity, Figure 17. This holds for all points on ∆, including the cusps, but also for the center C2. At poses where C reaches a cusp Pi′ of ∆, then for all lines Lθ∗(t) through P the point C∗ is a cusp of the corresponding envelope. Then the point is the so-called return pole, and the circular path of C the return circle or cuspidal circle [1, p. 274].

Corollary 8. The area of ∆θ∗ is independent of M and is given by:

c4 cos2 θ π A = . 2ab Note this is equal to cos2 θ of the area of ∆, see Equation 5.

Remark 11. For variable θ between 90◦ and 90◦, the said similarity defines an equiform motion where each point in− the plane runs along a circle through M with the same angular velocity. For each point, the configuration at θ =0 and M define a diameter of the trajectory. Recall the pre-images P of the cusps of ∆ have vertices at (t ), where t = u , i E i 1 − 3 t = u 2π , t = u 4π , see Theorem 2. 2 − 3 − 3 3 − 3 − 3 22 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

Corollary 9. The cusps P ∗ of ∆∗ have pre-images on which are invariant over i θ E θ and are congruent with the Pi,i =1, 2, 3.

Corollary 10. Lines PiPi∗ concur at C2∗.

Corollary 11. C∗ is a rotation of C by 2θ about the center X of . In particular, 2 2 3 K When θ = π/2, C2∗ = M, and ∆θ∗ degenerates to point M.

10. Conclusion Before we part, we would like to pay homage to eminent swiss mathematician Jakob Steiner (1796–1863), discoverer of several concepts appearing herein: the Steiner Ellipse and Inellipse, the Steiner Curve (or hypocycloid), the Steiner Point X99. Also due to him is the concept of orthologic triangles and the theorem of 3 concurrent osculating circles in the ellipse. Hats oﬀ and vielen dank, Herr Steiner! Some of the above phenomena are illustrated dynamically through the videos on Table 3.

PL# Title Narrated 01 Constant-Area Deltoid no 02 Properties of the Deltoid yes 03 Osculating Circles at the Cusp Pre-Images yes

04 Loci of Cusps and C2 no 05 Concyclic pre-images, osculating circles, no and 3 area-invariant triangles 06 Rotated Negative Pedal Curve yes

Table 3. Playlist of videos. Column “PL#” indicates the entry within the playlist. STEINER’S HAT: A CONSTANT-AREA DELTOID 23

Appendix Explicit Expressions for the A. Pi, Pi′, Pi′′

u u P = a cos , b sin , 1 3 − 3 h u π i u π P = a sin( + ), b cos( + ) 2 − 3 6 − 3 6 h u π u π i P = a cos( + ),b sin( + ) 3 − 3 6 3 6 h i 3c2 u a2 + b2 3c2 u a2 + b2 P1′ = cos cos u, sin sin u " 2a 3 − 2a 2b 3 − 2b # 3c2 u 3√3c2 u a2 + b2 3√3c2 u 3c2 u a2 + b2 P2′ = cos sin cos u, cos sin sin u "− 4a 3 − 4a 3 − 2a 4b 3 − 4b 3 − 2b # 3c2 u 3√3c2 u a2 + b2 3√3c2 u 3c2 u a2 + b2 P3′ = cos + sin cos u, cos sin sin u "− 4a 3 4a 3 − 2a − 4b 3 − 4b 3 − 2b # 3c2 u c2 3c2 u c2 P ′′ = cos + cos u, sin + sin u 1 4a 3 4a 4b 3 4b 3c2 u 3√3c2 u c2 3√3c2 u 3c2 u c2 P2′′ = cos sin + cos u, cos sin sin u "− 8a 3 − 8a 3 4a 8b 3 − 8b 3 − 4b # 3c2 u 3√3c2 u c2 3√3c2 u 3c2 u c2 P3′′ = cos + sin + cos u, cos sin sin u "− 8a 3 8a 3 4a − 8b 3 − 8b 3 − 4b #

Appendix B. Table of Symbols 24 R.GARCIA,D.REZNIK,H.STACHEL,ANDM.HELMAN

symbol meaning note main ellipse E a,b major, minor semi-axes of E c half the focal length of c2 = a2 b2 E − O center E M,M a ﬁxed point on the boundary of [a cos u,b sin u], u E perspector of T ′,T ′′, = X99 P (t) a point which sweeps the boundary of [a cos t,b sin t] E L(t) Line through P (t) perp. to P (t) M − ∆, ∆u Steiner’s Hat, negative pedal curve invariant area of with respect to M E ∆θ∗ envelope of L(t) rotated θ about P (t) invariant area ¯ C average coordinates of ∆ = C2 = X2′

C2, C2∗ area center of ∆, ∆∗ C2 = X2′ = X98

Pi′, Pi, Pi′′ The cusps of ∆, their pre-images, and centers of Ki (see below)

Pi∗ cusps of ∆∗ PiPi∗ concur at C2∗

T,T ′,T ′′ triangles deﬁned by the Pi, Pi′, Pi′′ invariant area over M A, A′, A′′ areas of T,T ′,T ′′ A′/A′′ =4 for any M,a,b Steiner’s Curve aka. Hypocycloid and Triscupoid S ′ Steiner Circumellipse centered at C E 2 of cusp (Pi′) triangle a′,b′ major, minor semi-axes of ′ invariant, axis-parallel E and similar to 90◦-rotated E Circumcircle of T center X , contains M, P , C , C∗ K 3 i 2 2 ′ Circumcircle of T ′ K Circles osculating at the P contain P , P ′,M Ki E i i i ∗ evolute of the K lie on it E E i Apollonius Hyperbola of wrt M ∆ is tangent to at H E E H∩E X3 circumcenter of T = X2′′ X4 perspector of T and T ′′

X99 Steiner Point of T = M

X98 Tarry Point of T = C2

X2′ centroid of T ′ = C2, and perspector of T,T ′

X99′ Steiner Point of T ′

X2′′ centroid of T ′′ = X3

Table 4. All Symbols used. STEINER’S HAT: A CONSTANT-AREA DELTOID 25

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